# simple graphs of degree 16 with a semiregular normal subgroup isomorphic to the quaternion group $Q_8$

Is there any simple graph $\Gamma$ with 16 vertices with full automorphism group $G$ such that $H\cong Q_8$ be a semiregular normal subgroup of $G$?

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$Q_8$ has outer automorphisms of order 2. So you can extend $Q_8$ using such an automorphism, obtaining $G$ of order 16... I hope it is not your homework. :) –  Dima Pasechnik Dec 25 '12 at 17:07
Thanks for your answer. Indeed I want to be $G$ as full automorphism group of a graph with 16 vertices. So I change my question. –  majid arezoomand Dec 25 '12 at 19:45

## 1 Answer

Here is an example.

Adjacency matrix:

$\left[ \begin{array}{cccccccccccccccc} 0&1&1&0&1&1&1&0&0&0&0&0&0&0&0&0\\1&0&0&1&1&1&0&1&0&0&0&0&0&0&0&0\\1&0&0&0&0&0&1&0&1&1&1&0&0&0&0&0\\0&1&0&0&0&0&0&1&1&1&0&1&0&0&0&0\\1&1&0&0&0&0&0&0&0&0&1&0&1&0&1&0\\1&1&0&0&0&0&0&0&0&0&0&1&0&1&0&1\\1&0&1&0&0&0&0&0&0&1&0&0&1&1&0&0\\0&1&0&1&0&0&0&0&1&0&0&0&1&1&0&0\\0&0&1&1&0&0&0&1&0&0&1&0&0&0&0&1\\0&0&1&1&0&0&1&0&0&0&0&1&0&0&1&0\\0&0&1&0&1&0&0&0&1&0&0&0&0&0&1&1\\0&0&0&1&0&1&0&0&0&1&0&0&0&0&1&1\\0&0&0&0&1&0&1&1&0&0&0&0&0&1&1&0\\0&0&0&0&0&1&1&1&0&0&0&0&1&0&0&1\\0&0&0&0&1&0&0&0&0&1&1&1&1&0&0&0\\0&0&0&0&0&1&0&0&1&0&1&1&0&1&0&0 \end{array} \right]$

Neighbours of each vertex:

$\begin{array}{c|ccccc} 1&2&3&5&6&7\\ 2&1&4&5&6&8\\ 3&1&7&9&10&11\\ 4&2&8&9&10&12\\ 5&1&2&11&13&15\\ 6&1&2&12&14&16\\ 7&1&3&10&13&14\\ 8&2&4&9&13&14\\ 9&3&4&8&11&16\\ 10&3&4&7&12&15\\ 11&3&5&9&15&16\\ 12&4&6&10&15&16\\ 13&5&7&8&14&15\\ 14&6&7&8&13&16\\ 15&5&10&11&12&13\\ 16&6&9&11&12&14 \end{array}$

It is a Cayley graph on the semidihedral group of order 16. In fact, it is a graphical regular representation, so the full automorphism group acts regularly on the vertices. Note that the semidihedral group of order 16 has a (normal) subgroup isomorphic to the quaternion group.

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